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                <a class="post-title-link" href="/2017/03/01/247/" itemprop="url">2.28模拟题 problem1 判断三角形</a></h1>
        

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            <p>题目描述<br>Shoutmon给萌萌兽出了一个题：给出三条线段的长度，问能否组成三角形。不过线段的长度是通过每一个数位的英文给出的。你能帮助萌萌兽吗？</p>
<p>输入<br>每个输入文件中一组数据。</p>
<p>分三行分别给出三条线段的长度。每行以一个整数N（1&lt;=N&lt;=5）开始，表示后面给出的单词个数；接下来跟着N个英文单词，每个单词表示一个数位，线段的长度由这些英文单词直接拼接而成（例如one two three代表123，one one代表11），每两个单词之间用一个空格隔开。数字0到9分别用下面十个单词表示：zero、one、two、three、four、five、six、seven、eight、nine。数据保证每条线段的长度都不为0。</p>
<p>输出<br>如果可以组成三角形，那么输出YES；否则输出NO。</p>
<p>样例输入<br>1 one<br>2 one one<br>3 one one one<br>样例输出<br>NO<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br></pre></td><td class="code"><pre><span class="line">#include &lt;iostream&gt;</span><br><span class="line">#include &lt;stdio.h&gt;</span><br><span class="line">#include &lt;algorithm&gt;</span><br><span class="line">#include &lt;string&gt;</span><br><span class="line">#include &lt;vector&gt;</span><br><span class="line">#include &lt;queue&gt;</span><br><span class="line">#include &lt;map&gt;</span><br><span class="line">using namespace std;</span><br><span class="line">string numcode[] = &#123;&quot;zero&quot;, &quot;one&quot;, &quot;two&quot;, &quot;three&quot;, &quot;four&quot;, &quot;five&quot;, &quot;six&quot;, &quot;seven&quot;, &quot;eight&quot;, &quot;nine&quot;&#125;;</span><br><span class="line">map&lt;string, int&gt; strToNum;</span><br><span class="line">int main() &#123;</span><br><span class="line">    int tri[3];</span><br><span class="line">    for (int i = 0; i &lt; 10; i++) &#123;</span><br><span class="line">        strToNum[numcode[i]] = i;</span><br><span class="line">    &#125;</span><br><span class="line">    for (int i = 0; i &lt; 3; i++) &#123;</span><br><span class="line">        int k, number = 0;</span><br><span class="line">        scanf(&quot;%d&quot;, &amp;k);</span><br><span class="line">        for (int j = 0; j &lt; k; j++) &#123;</span><br><span class="line">            char temp[10];</span><br><span class="line">            scanf(&quot;%s&quot;, temp);</span><br><span class="line">            number = number * 10 + strToNum[temp];</span><br><span class="line">        &#125;</span><br><span class="line">        tri[i] = number;</span><br><span class="line">    &#125;</span><br><span class="line">    if(tri[0] + tri[1] &gt; tri[2] &amp;&amp;</span><br><span class="line">       tri[1] + tri[2] &gt; tri[0] &amp;&amp;</span><br><span class="line">       tri[0] + tri[2] &gt; tri[1] )&#123;</span><br><span class="line">        printf(&quot;YES\n&quot;);</span><br><span class="line">    &#125;else&#123;</span><br><span class="line">        printf(&quot;NO\n&quot;);</span><br><span class="line">    &#125;</span><br><span class="line">    return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>

          
        
      
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                <a class="post-title-link" href="/2017/03/01/245/" itemprop="url">2.28模拟题 problem3 还原二叉树</a></h1>
        

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            <p>题目描述<br>给一棵二叉树的层序遍历序列和中序遍历序列，求这棵二叉树的先序遍历序列和后序遍历序列。</p>
<p>输入<br>每个输入文件中一组数据。</p>
<p>第一行一个正整数N（1&lt;=N&lt;=30），代表二叉树的结点个数（结点编号为1~N）。接下来两行，每行N个正整数，分别代表二叉树的层序遍历序列和中序遍历序列。数据保证序列中1~N的每个数出现且只出现一次。</p>
<p>输出<br>输出两行，每行N个正整数，分别代表二叉树的先序遍历序列和后序遍历序列。每行末尾不输出额外的空格。</p>
<p>样例输入<br>7<br>3 5 4 2 6 7 1<br>2 5 3 6 4 7 1<br>样例输出<br>3 5 2 4 6 7 1<br>2 5 6 1 7 4 3<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br><span class="line">73</span><br><span class="line">74</span><br><span class="line">75</span><br><span class="line">76</span><br><span class="line">77</span><br><span class="line">78</span><br><span class="line">79</span><br><span class="line">80</span><br><span class="line">81</span><br><span class="line">82</span><br><span class="line">83</span><br><span class="line">84</span><br><span class="line">85</span><br><span class="line">86</span><br><span class="line">87</span><br></pre></td><td class="code"><pre><span class="line">#include &lt;iostream&gt;</span><br><span class="line">#include &lt;stdio.h&gt;</span><br><span class="line">#include &lt;algorithm&gt;</span><br><span class="line">#include &lt;string&gt;</span><br><span class="line">#include &lt;vector&gt;</span><br><span class="line">#include &lt;queue&gt;</span><br><span class="line">using namespace std;</span><br><span class="line">const int maxn = 50;</span><br><span class="line">struct node&#123;</span><br><span class="line">    int data;</span><br><span class="line">    node *lchild, *rchild;</span><br><span class="line">&#125;;</span><br><span class="line">int n, lev[maxn], in[maxn];</span><br><span class="line">vector&lt;int&gt; layer, pre, post;</span><br><span class="line">node* createTree(vector&lt;int&gt; layer, int inL, int inR)&#123;</span><br><span class="line">    if (layer.size() == 0) &#123;</span><br><span class="line">        return NULL;</span><br><span class="line">    &#125;</span><br><span class="line">    node* root = new node;</span><br><span class="line">    root-&gt;data = layer[0];</span><br><span class="line">    int k;</span><br><span class="line">    for (k = inL; k &lt;= inR; k++) &#123;</span><br><span class="line">        if (layer[0] == in[k]) &#123;//在中序遍历中找到</span><br><span class="line">            break;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    vector&lt;int&gt; layerLeft;</span><br><span class="line">    vector&lt;int&gt; layerRight;</span><br><span class="line">    for (int i = 1; i &lt; layer.size(); i++) &#123;</span><br><span class="line">        bool isLeft = false;</span><br><span class="line">        for (int j = inL; j &lt; k; j++) &#123;</span><br><span class="line">            if (layer[i] == in[j]) &#123;</span><br><span class="line">                isLeft = true;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        if (isLeft) &#123;</span><br><span class="line">            layerLeft.push_back(layer[i]);</span><br><span class="line">        &#125;else&#123;</span><br><span class="line">            layerRight.push_back(layer[i]);</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    root-&gt;lchild = createTree(layerLeft, inL, k - 1);</span><br><span class="line">    root-&gt;rchild = createTree(layerRight, k + 1, inR);</span><br><span class="line">    return root;</span><br><span class="line">&#125;</span><br><span class="line">void preOrder(node* root, vector&lt;int&gt; &amp;vi) &#123;</span><br><span class="line">    if (root == NULL) &#123;</span><br><span class="line">        return;</span><br><span class="line">    &#125;</span><br><span class="line">    vi.push_back(root-&gt;data);</span><br><span class="line">    preOrder(root-&gt;lchild, vi);</span><br><span class="line">    preOrder(root-&gt;rchild, vi);</span><br><span class="line">&#125;</span><br><span class="line">void postOrder(node* root, vector&lt;int&gt; &amp;vi) &#123;</span><br><span class="line">    if (root == NULL) &#123;</span><br><span class="line">        return;</span><br><span class="line">    &#125;</span><br><span class="line">    postOrder(root-&gt;lchild, vi);</span><br><span class="line">    postOrder(root-&gt;rchild, vi);</span><br><span class="line">    vi.push_back(root-&gt;data);</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">int main() &#123;</span><br><span class="line">    scanf(&quot;%d&quot;, &amp;n);</span><br><span class="line">    for (int i = 0; i &lt; n; i++)&#123;</span><br><span class="line">        scanf(&quot;%d&quot;, &amp;lev[i]);</span><br><span class="line">        layer.push_back(lev[i]);</span><br><span class="line">    &#125;</span><br><span class="line">    for (int i = 0; i &lt; n; i++) &#123;</span><br><span class="line">        scanf(&quot;%d&quot;, &amp;in[i]);</span><br><span class="line">    &#125;</span><br><span class="line">    node* root = NULL;</span><br><span class="line">    root = createTree(layer, 0, n - 1);</span><br><span class="line">    preOrder(root, pre);</span><br><span class="line">    postOrder(root, post);</span><br><span class="line">    for (int i = 0; i &lt; n; i++) &#123;</span><br><span class="line">        printf(&quot;%d&quot;, pre[i]);</span><br><span class="line">        if (i &lt; n - 1) printf(&quot; &quot;);</span><br><span class="line">        else printf(&quot;\n&quot;);</span><br><span class="line">    &#125;</span><br><span class="line">    for (int i = 0; i &lt; n; i++) &#123;</span><br><span class="line">        printf(&quot;%d&quot;, post[i]);</span><br><span class="line">        if (i &lt; n - 1) printf(&quot; &quot;);</span><br><span class="line">        else printf(&quot;\n&quot;);</span><br><span class="line">    &#125;</span><br><span class="line">    return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>

          
        
      
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                <a class="post-title-link" href="/2017/03/01/244/" itemprop="url">2.28模拟题 problem2 进击的二叉查找树</a></h1>
        

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            <p>题目描述<br>给定1~N的两个排列，使用这两个排列分别构建两棵二叉查找树（也就是通过往一棵空树中依次插入序列元素的构建方式）。如果这两棵二叉查找树完全相同，那么输出YES；否则输出NO。之后，输出第一个排列对应的二叉查找树的后序序列、层序序列。</p>
<p>输入<br>每个输入文件中一组数据。</p>
<p>第一行1个正整数N（1&lt;=N&lt;=30），表示二叉查找树中的结点个数。</p>
<p>接下来两行，代表1~N的两个排列。</p>
<p>输出<br>如果两个排列代表的二叉查找树完全相同，那么输出一行YES，否则输出一行NO。</p>
<p>接下来两行分别输出第一个排列对应的二叉查找树的后序序列、层序序列，整数之间用空格隔开。</p>
<p>每行末尾不允许有多余的空格。</p>
<p>样例输入<br>5<br>4 2 1 3 5<br>4 5 2 3 1<br>样例输出<br>YES<br>1 3 2 5 4<br>4 2 5 1 3<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br><span class="line">73</span><br><span class="line">74</span><br><span class="line">75</span><br><span class="line">76</span><br><span class="line">77</span><br><span class="line">78</span><br><span class="line">79</span><br><span class="line">80</span><br><span class="line">81</span><br><span class="line">82</span><br><span class="line">83</span><br><span class="line">84</span><br><span class="line">85</span><br><span class="line">86</span><br><span class="line">87</span><br><span class="line">88</span><br><span class="line">89</span><br><span class="line">90</span><br><span class="line">91</span><br><span class="line">92</span><br></pre></td><td class="code"><pre><span class="line">#include &lt;iostream&gt;</span><br><span class="line">#include &lt;stdio.h&gt;</span><br><span class="line">#include &lt;algorithm&gt;</span><br><span class="line">#include &lt;string&gt;</span><br><span class="line">#include &lt;vector&gt;</span><br><span class="line">#include &lt;queue&gt;</span><br><span class="line">using namespace std;</span><br><span class="line">const int maxn = 40;</span><br><span class="line">struct node&#123;</span><br><span class="line">    int data;</span><br><span class="line">    int lchild;</span><br><span class="line">    int rchild;</span><br><span class="line">&#125;Node[maxn * 2];</span><br><span class="line">int nodenum = 0;</span><br><span class="line">int n;</span><br><span class="line">vector&lt;int&gt; post1, post2, lev1;</span><br><span class="line">int newNode(int x)&#123;</span><br><span class="line">    Node[nodenum].data = x;</span><br><span class="line">    Node[nodenum].lchild = -1;</span><br><span class="line">    Node[nodenum].rchild = -1;</span><br><span class="line">    return nodenum++;</span><br><span class="line">&#125;</span><br><span class="line">void insert(int &amp;root, int x)&#123;</span><br><span class="line">    if (root == -1) &#123;</span><br><span class="line">        root = newNode(x);</span><br><span class="line">        return;</span><br><span class="line">    &#125;</span><br><span class="line">    if (x &lt; Node[root].data) &#123;</span><br><span class="line">        insert(Node[root].lchild, x);</span><br><span class="line">    &#125;</span><br><span class="line">    if (x &gt; Node[root].data) &#123;</span><br><span class="line">        insert(Node[root].rchild, x);</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br><span class="line">int createTree(int arr[])&#123;</span><br><span class="line">    int root = -1;</span><br><span class="line">    for (int i = 0; i &lt; n; i++) &#123;</span><br><span class="line">        insert(root, arr[i]);</span><br><span class="line">    &#125;</span><br><span class="line">    return root;</span><br><span class="line">&#125;</span><br><span class="line">void post(int root, vector&lt;int&gt;&amp; vi)&#123;</span><br><span class="line">    if (root == -1) &#123;</span><br><span class="line">        return;</span><br><span class="line">    &#125;</span><br><span class="line">    post(Node[root].lchild, vi);</span><br><span class="line">    post(Node[root].rchild, vi);</span><br><span class="line">    vi.push_back(Node[root].data);</span><br><span class="line">&#125;</span><br><span class="line">void lev(int root, vector&lt;int&gt;&amp; vi)&#123;</span><br><span class="line">    queue&lt;int&gt; q;</span><br><span class="line">    q.push(root);</span><br><span class="line">    while (!q.empty()) &#123;</span><br><span class="line">        int now = q.front();</span><br><span class="line">        q.pop();</span><br><span class="line">        vi.push_back(Node[now].data);</span><br><span class="line">        if (Node[now].lchild != -1) q.push(Node[now].lchild);</span><br><span class="line">        if (Node[now].rchild != -1) q.push(Node[now].rchild);</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br><span class="line">int main() &#123;</span><br><span class="line">    int str1[maxn], str2[maxn];</span><br><span class="line">    scanf(&quot;%d&quot;, &amp;n);</span><br><span class="line">    for (int i = 0; i &lt; n; i++) &#123;</span><br><span class="line">        scanf(&quot;%d&quot;, &amp;str1[i]);</span><br><span class="line">    &#125;</span><br><span class="line">    int root1 = createTree(str1);</span><br><span class="line">    for (int i = 0; i &lt; n; i++) &#123;</span><br><span class="line">        scanf(&quot;%d&quot;, &amp;str2[i]);</span><br><span class="line">    &#125;</span><br><span class="line">    int root2 = createTree(str2);</span><br><span class="line">    post(root1, post1);</span><br><span class="line">    post(root2, post2);</span><br><span class="line">    lev(root1, lev1);</span><br><span class="line">    if (post1 == post2) &#123;</span><br><span class="line">        printf(&quot;YES\n&quot;);</span><br><span class="line">    &#125;else&#123;</span><br><span class="line">        printf(&quot;NO\n&quot;);</span><br><span class="line">    &#125;</span><br><span class="line">    for (int i = 0; i &lt; n; i++) &#123;</span><br><span class="line">        printf(&quot;%d&quot;, post1[i]);</span><br><span class="line">        if(i &lt; n - 1)printf(&quot; &quot;);</span><br><span class="line">        else printf(&quot;\n&quot;);</span><br><span class="line">    &#125;</span><br><span class="line">    for (int i = 0; i &lt; n; i++) &#123;</span><br><span class="line">        printf(&quot;%d&quot;, lev1[i]);</span><br><span class="line">        if(i &lt; n - 1)printf(&quot; &quot;);</span><br><span class="line">        else printf(&quot;\n&quot;);</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">    return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>

          
        
      
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                <a class="post-title-link" href="/2017/02/28/243/" itemprop="url">PAT A1040 . Longest Symmetric String (25)</a></h1>
        

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            <p>Given a string, you are supposed to output the length of the longest symmetric sub-string. For example, given “Is PAT&amp;TAP symmetric?”, the longest symmetric sub-string is “s PAT&amp;TAP s”, hence you must output 11.</p>
<p>Input Specification:</p>
<p>Each input file contains one test case which gives a non-empty string of length no more than 1000.</p>
<p>Output Specification:</p>
<p>For each test case, simply print the maximum length in a line.</p>
<p>Sample Input:<br>Is PAT&amp;TAP symmetric?<br>Sample Output:<br>11<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br></pre></td><td class="code"><pre><span class="line">#include &quot;stdio.h&quot;</span><br><span class="line">//#include &quot;math.h&quot;</span><br><span class="line">#include &quot;string.h&quot;</span><br><span class="line">//#include &quot;iostream&quot;</span><br><span class="line">//#include &quot;stdlib.h&quot;</span><br><span class="line">//#include &quot;vector&quot;</span><br><span class="line">//#include &quot;set&quot;</span><br><span class="line">//#include &quot;map&quot;</span><br><span class="line">//#include &quot;stack&quot;</span><br><span class="line">//#include &quot;queue&quot;</span><br><span class="line">#include &quot;algorithm&quot;</span><br><span class="line">using namespace std;</span><br><span class="line"></span><br><span class="line">const int maxn = 1010;</span><br><span class="line">char S[maxn];</span><br><span class="line">int dp[maxn][maxn];</span><br><span class="line"></span><br><span class="line">int main()&#123;</span><br><span class="line">    gets(S);</span><br><span class="line">    int len = strlen(S), ans = 1;</span><br><span class="line">    memset(dp, 0, sizeof(S));</span><br><span class="line">    //边界</span><br><span class="line">    for (int i = 0; i &lt; len; i++) &#123;</span><br><span class="line">        dp[i][i] = 1;</span><br><span class="line">        if (i &lt; len - 1) &#123;</span><br><span class="line">            if (S[i] == S[i + 1]) &#123;</span><br><span class="line">                dp[i][i + 1] = 1;</span><br><span class="line">                ans = 2;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    //状态转移方程</span><br><span class="line">    for (int L = 3; L &lt;= len; L++) &#123;</span><br><span class="line">        for (int i = 0; i + L - 1 &lt; len; i++) &#123;//枚举子串的起始结点</span><br><span class="line">            int j = i + L - 1;//子串右结点</span><br><span class="line">            if (S[i] == S[j] &amp;&amp; dp[i + 1][j - 1] == 1) &#123;</span><br><span class="line">                dp[i][j] = 1;</span><br><span class="line">                ans = L;//更新最长回文子串长度</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    printf(&quot;%d\n&quot;, ans);</span><br><span class="line">    return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>

          
        
      
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                <a class="post-title-link" href="/2017/02/27/240/" itemprop="url">PAT A1045 . Favorite Color Stripe (30)</a></h1>
        

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            <p>Eva is trying to make her own color stripe out of a given one. She would like to keep only her favorite colors in her favorite order by cutting off those unwanted pieces and sewing the remaining parts together to form her favorite color stripe.</p>
<p>It is said that a normal human eye can distinguish about less than 200 different colors, so Eva’s favorite colors are limited. However the original stripe could be very long, and Eva would like to have the remaining favorite stripe with the maximum length. So she needs your help to find her the best result.</p>
<p>Note that the solution might not be unique, but you only have to tell her the maximum length. For example, given a stripe of colors {2 2 4 1 5 5 6 3 1 1 5 6}. If Eva’s favorite colors are given in her favorite order as {2 3 1 5 6}, then she has 4 possible best solutions {2 2 1 1 1 5 6}, {2 2 1 5 5 5 6}, {2 2 1 5 5 6 6}, and {2 2 3 1 1 5 6}.</p>
<p>Input Specification:</p>
<p>Each input file contains one test case. For each case, the first line contains a positive integer N (&lt;=200) which is the total number of colors involved (and hence the colors are numbered from 1 to N). Then the next line starts with a positive integer M (&lt;=200) followed by M Eva’s favorite color numbers given in her favorite order. Finally the third line starts with a positive integer L (&lt;=10000) which is the length of the given stripe, followed by L colors on the stripe. All the numbers in a line are separated by a space.</p>
<p>Output Specification:</p>
<p>For each test case, simply print in a line the maximum length of Eva’s favorite stripe.</p>
<p>Sample Input:<br>6<br>5 2 3 1 5 6<br>12 2 2 4 1 5 5 6 3 1 1 5 6<br>Sample Output:<br>7</p>
<p>lis:<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br></pre></td><td class="code"><pre><span class="line">#include &quot;stdio.h&quot;</span><br><span class="line">#include &quot;algorithm&quot;</span><br><span class="line">using namespace std;</span><br><span class="line"></span><br><span class="line">const int maxc = 210;//最大颜色数</span><br><span class="line">const int maxn = 10010;//最大L</span><br><span class="line">int hashTable[maxc];//将最喜欢的颜色序列映射为递增序列，不喜欢的颜色映射为-1</span><br><span class="line">int A[maxn], dp[maxn];</span><br><span class="line">int main()&#123;</span><br><span class="line">    int n, m, x;</span><br><span class="line">    scanf(&quot;%d%d&quot;, &amp;n, &amp;m);</span><br><span class="line">    memset(hashTable, -1, sizeof(hashTable));</span><br><span class="line">    for (int i = 0; i &lt; m; i++) &#123;</span><br><span class="line">        scanf(&quot;%d&quot;, &amp;x);</span><br><span class="line">        hashTable[x] = i;</span><br><span class="line">    &#125;</span><br><span class="line">    int L, num = 0;//num存放颜色序列中eva最喜欢的颜色的总数</span><br><span class="line">    scanf(&quot;%d&quot;, &amp;L);</span><br><span class="line">    for (int i = 0; i &lt; L; i++) &#123;</span><br><span class="line">        scanf(&quot;%d&quot;, &amp;x);</span><br><span class="line">        if (hashTable[x] &gt;= 0) &#123;//若有喜欢的颜色，添加到数组A中</span><br><span class="line">            A[num++] = hashTable[x];</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    //以下全部为LIS问题模板</span><br><span class="line">    int ans = -1;</span><br><span class="line">    for (int i = 0; i &lt; num; i++) &#123;</span><br><span class="line">        dp[i] = 1;</span><br><span class="line">        for (int j = 0; j &lt; i; j++) &#123;</span><br><span class="line">            if (A[j] &lt;= A[i] &amp;&amp; dp[i] &lt; dp[j] + 1) &#123;</span><br><span class="line">                dp[i] = dp[j] + 1;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        ans = max(ans, dp[i]);</span><br><span class="line">    &#125;</span><br><span class="line">    printf(&quot;%d&quot;, ans);</span><br><span class="line">    return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>
<p>lcs:</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br></pre></td><td class="code"><pre><span class="line">#include &quot;stdio.h&quot;</span><br><span class="line">#include &quot;math.h&quot;</span><br><span class="line">#include &quot;string.h&quot;</span><br><span class="line"></span><br><span class="line">const int maxc = 210;</span><br><span class="line">const int maxn = 1000010;</span><br><span class="line">int a[maxc], b[maxn], dp[maxc][maxn];</span><br><span class="line">int main()&#123;</span><br><span class="line">    int n, m;</span><br><span class="line">    scanf(&quot;%d%d&quot;, &amp;n, &amp;m);</span><br><span class="line">    for (int i = 1; i &lt;= m ; i++) &#123;</span><br><span class="line">        scanf(&quot;%d&quot;, &amp;a[i]);//读入序列A</span><br><span class="line">    &#125;</span><br><span class="line">    int L;</span><br><span class="line">    scanf(&quot;%d&quot;, &amp;L);</span><br><span class="line">    for (int i = 1; i &lt;= L; i++) &#123;</span><br><span class="line">        scanf(&quot;%d&quot;, &amp;b[i]);</span><br><span class="line">    &#125;</span><br><span class="line">    //边界</span><br><span class="line">    for (int i = 0; i &lt;= m; i++) &#123;</span><br><span class="line">        dp[i][0] = 0;</span><br><span class="line">    &#125;</span><br><span class="line">    for (int j = 0; j &lt;= L; j++) &#123;</span><br><span class="line">        dp[0][j] = 0;</span><br><span class="line">    &#125;</span><br><span class="line">    //状态转移方程</span><br><span class="line">    for (int i = 1; i &lt;= m; i++) &#123;</span><br><span class="line">        for (int j = 1; j &lt;= L; j++) &#123;</span><br><span class="line">            //取dp[i-1][j]、dp[i][j-1]中的较大值</span><br><span class="line">            int MAX = max(dp[i - 1][j], dp[i][j - 1]);</span><br><span class="line">            if (a[i] == b[j]) &#123;</span><br><span class="line">                dp[i][j] = MAX + 1;</span><br><span class="line">            &#125;else&#123;</span><br><span class="line">                dp[i][j] = MAX;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    //输出答案</span><br><span class="line">    printf(&quot;%d\n&quot;, dp[m][L]);</span><br><span class="line">    return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
          
        
      
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                <a class="post-title-link" href="/2017/02/27/238/" itemprop="url">PAT A1007 . Maximum Subsequence Sum (25)</a></h1>
        

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            <p>Given a sequence of K integers { N<sub>1</sub>, N<sub>2</sub>, …, N<sub>K</sub> }. A continuous subsequence is defined to be { N<sub>i</sub>, N<sub>i+1</sub>, …, N<sub>j</sub> } where 1 &lt;= i &lt;= j &lt;= K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.</p>
<p>Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.</p>
<p>Input Specification:</p>
<p>Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (&lt;= 10000). The second line contains K numbers, separated by a space.</p>
<p>Output Specification:</p>
<p>For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.</p>
<p>Sample Input:<br>10<br>-10 1 2 3 4 -5 -23 3 7 -21<br>Sample Output:<br>10 1 4<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br></pre></td><td class="code"><pre><span class="line">#include &quot;stdio.h&quot;</span><br><span class="line"></span><br><span class="line">const int maxn = 10010;</span><br><span class="line">int a[maxn], dp[maxn];</span><br><span class="line">int s[maxn] = &#123;0&#125;;//s[i]表示dp[i]的连续序列从a的哪一个元素开始</span><br><span class="line">    </span><br><span class="line">int main()&#123;</span><br><span class="line">    int n;</span><br><span class="line">    scanf(&quot;%d&quot;, &amp;n);</span><br><span class="line">    bool flag = false;//flag表示数组a中是否全小于0</span><br><span class="line">    for(int i = 0; i &lt; n; i++)&#123;</span><br><span class="line">        scanf(&quot;%d&quot;, &amp;a[i]);</span><br><span class="line">        if (a[i] &gt;= 0) flag = true;</span><br><span class="line">    &#125;</span><br><span class="line">    if (flag == false) &#123;//如果所有数字都小于0，则输出0以及尾首元素</span><br><span class="line">        printf(&quot;0 %d %d\n&quot;, a[0], a[n - 1]);</span><br><span class="line">        return 0;</span><br><span class="line">    &#125;</span><br><span class="line">    //边界</span><br><span class="line">    dp[0] = a[0];</span><br><span class="line">    for (int i = 1; i &lt; n; i++) &#123;</span><br><span class="line">        //状态转移方程</span><br><span class="line">        if (dp[i - 1] + a[i] &gt; a[i]) &#123;</span><br><span class="line">            dp[i] = dp[i - 1] + a[i];</span><br><span class="line">            s[i] = s[i - 1];</span><br><span class="line">        &#125;else&#123;</span><br><span class="line">            dp[i] = a[i];</span><br><span class="line">            s[i] = i;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    //因为dp[i]存放的是以a[i]结尾的连续序列的最大和</span><br><span class="line">    //因此需要遍历i得到最大的才是结果</span><br><span class="line">    int k = 0;</span><br><span class="line">    for (int i = 1; i &lt; n; i++) &#123;</span><br><span class="line">        if (dp[i] &gt; dp[k]) &#123;</span><br><span class="line">            k = i;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    printf(&quot;%d %d %d\n&quot;, dp[k], a[s[k]], a[k]);</span><br><span class="line">    return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>

          
        
      
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                <a class="post-title-link" href="/2017/02/26/237/" itemprop="url">PAT A1087</a></h1>
        

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            <p>Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness.</p>
<p>Input Specification:</p>
<p>Each input file contains one test case. For each case, the first line contains 2 positive integers N (2&lt;=N&lt;=200), the number of cities, and K, the total number of routes between pairs of cities; followed by the name of the starting city. The next N-1 lines each gives the name of a city and an integer that represents the happiness one can gain from that city, except the starting city. Then K lines follow, each describes a route between two cities in the format “City1 City2 Cost”. Here the name of a city is a string of 3 capital English letters, and the destination is always ROM which represents Rome.</p>
<p>Output Specification:</p>
<p>For each test case, we are supposed to find the route with the least cost. If such a route is not unique, the one with the maximum happiness will be recommended. If such a route is still not unique, then we output the one with the maximum average happiness – it is guaranteed by the judge that such a solution exists and is unique.</p>
<p>Hence in the first line of output, you must print 4 numbers: the number of different routes with the least cost, the cost, the happiness, and the average happiness (take the integer part only) of the recommended route. Then in the next line, you are supposed to print the route in the format “City1-&gt;City2-&gt;…-&gt;ROM”.</p>
<p>Sample Input:<br>6 7 HZH<br>ROM 100<br>PKN 40<br>GDN 55<br>PRS 95<br>BLN 80<br>ROM GDN 1<br>BLN ROM 1<br>HZH PKN 1<br>PRS ROM 2<br>BLN HZH 2<br>PKN GDN 1<br>HZH PRS 1<br>Sample Output:<br>3 3 195 97<br>HZH-&gt;PRS-&gt;ROM</p>
<p>仅用dijstra：<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br><span class="line">73</span><br><span class="line">74</span><br><span class="line">75</span><br><span class="line">76</span><br><span class="line">77</span><br><span class="line">78</span><br><span class="line">79</span><br><span class="line">80</span><br><span class="line">81</span><br><span class="line">82</span><br><span class="line">83</span><br><span class="line">84</span><br><span class="line">85</span><br><span class="line">86</span><br><span class="line">87</span><br><span class="line">88</span><br><span class="line">89</span><br><span class="line">90</span><br><span class="line">91</span><br><span class="line">92</span><br><span class="line">93</span><br><span class="line">94</span><br><span class="line">95</span><br><span class="line">96</span><br><span class="line">97</span><br><span class="line">98</span><br><span class="line">99</span><br><span class="line">100</span><br><span class="line">101</span><br></pre></td><td class="code"><pre><span class="line">#include &quot;stdio.h&quot;</span><br><span class="line">#include &quot;math.h&quot;</span><br><span class="line">#include &quot;string.h&quot;</span><br><span class="line">#include &quot;iostream&quot;</span><br><span class="line">//#include &quot;stdlib.h&quot;</span><br><span class="line">#include &quot;vector&quot;</span><br><span class="line">#include &quot;set&quot;</span><br><span class="line">#include &quot;map&quot;</span><br><span class="line">//#include &quot;stack&quot;</span><br><span class="line">#include &quot;queue&quot;</span><br><span class="line">#include &quot;algorithm&quot;</span><br><span class="line">using namespace std;</span><br><span class="line"></span><br><span class="line">const int MAXV = 410;//最大顶点数</span><br><span class="line">const int INF = 1000000000;//无穷大</span><br><span class="line">//n 城市数目, k 边数, G 邻接矩阵, w 点权</span><br><span class="line">//d[]最短距离,dw[]最大点权,num[]最短路径条数</span><br><span class="line">//pt[]记录最短路径上的顶点数,pre[]记录前驱</span><br><span class="line">int n, k, G[MAXV][MAXV], weight[MAXV];</span><br><span class="line">int d[MAXV], w[MAXV], num[MAXV], pt[MAXV], pre[MAXV];</span><br><span class="line">bool vis[MAXV] = &#123;false&#125;;</span><br><span class="line">map&lt;string, int&gt; cityToIndex;</span><br><span class="line">map&lt;int, string&gt; indexToCity;</span><br><span class="line"></span><br><span class="line">void Dijkstra(int s)&#123;</span><br><span class="line">    fill(d, d + MAXV, INF);</span><br><span class="line">    memset(w, 0, sizeof(w));</span><br><span class="line">    memset(num, 0, sizeof(num));</span><br><span class="line">    memset(pt, 0, sizeof(pt));</span><br><span class="line">    for (int i = 0; i &lt; n; i++) pre[i] = i;//前驱初始化</span><br><span class="line">    d[s] = 0;</span><br><span class="line">    w[s] = 0;</span><br><span class="line">    num[s] = 1;</span><br><span class="line">    for (int i = 0; i &lt; n; i++) &#123;//循环n次</span><br><span class="line">        int u = -1, MIN = INF;</span><br><span class="line">        for (int j = 0; j &lt; n; j++) &#123;</span><br><span class="line">            if (vis[j] == false &amp;&amp; d[j] &lt; MIN) &#123;</span><br><span class="line">                u = j;</span><br><span class="line">                MIN = d[j];</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        if (u == -1) return;</span><br><span class="line">        vis[u] = true;</span><br><span class="line">        for (int v = 0; v &lt; n; v++) &#123;</span><br><span class="line">            if (vis[v] == false &amp;&amp; G[u][v] != INF) &#123;</span><br><span class="line">                if (d[u] + G[u][v] &lt; d[v]) &#123;</span><br><span class="line">                    d[v] = d[u] + G[u][v];//优化d[v]</span><br><span class="line">                    w[v] = w[u] + weight[v];</span><br><span class="line">                    num[v] = num[u];</span><br><span class="line">                    pt[v] = pt[u] + 1;</span><br><span class="line">                    pre[v] = u;</span><br><span class="line">                &#125;else if (d[u] + G[u][v] == d[v])&#123;</span><br><span class="line">                    num[v] += num[u];//加最短路径条数</span><br><span class="line">                    if (w[u] + weight[v] &gt; w[v]) &#123;</span><br><span class="line">                        w[v] = w[u] + weight[v];</span><br><span class="line">                        pt[v] = pt[u] + 1;</span><br><span class="line">                        pre[v] = u;</span><br><span class="line">                    &#125;else if(w[u] + weight[v] == w[v])&#123;</span><br><span class="line">                        double uAvg = 1.0 * (w[u] + weight[v]) / (pt[u] + 1);</span><br><span class="line">                        double vAvg = 1.0 * w[v] / pt[v];</span><br><span class="line">                        if (uAvg &gt; vAvg) &#123;</span><br><span class="line">                            pt[v] = pt[u] + 1;</span><br><span class="line">                            pre[v] = u;</span><br><span class="line">                        &#125;</span><br><span class="line">                    &#125;</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br><span class="line">void printPath(int v)&#123;</span><br><span class="line">    if (v == 0) &#123;</span><br><span class="line">        cout &lt;&lt; indexToCity[v];</span><br><span class="line">        return;</span><br><span class="line">    &#125;</span><br><span class="line">    printPath(pre[v]);</span><br><span class="line">    cout &lt;&lt;&quot;-&gt;&quot;&lt;&lt;indexToCity[v];</span><br><span class="line">&#125;</span><br><span class="line">int main()&#123;</span><br><span class="line">    string start, str1, str2;</span><br><span class="line">    cin &gt;&gt; n &gt;&gt; k &gt;&gt; start;</span><br><span class="line">    cityToIndex[start] = 0;</span><br><span class="line">    indexToCity[0] = start;</span><br><span class="line">    for (int i = 1; i &lt;= n - 1; i++) &#123;</span><br><span class="line">        cin &gt;&gt; str1 &gt;&gt; weight[i];//读入happy值，点权</span><br><span class="line">        cityToIndex[str1] = i;</span><br><span class="line">        indexToCity[i] = str1;</span><br><span class="line">    &#125;</span><br><span class="line">    fill(G[0], G[0] + MAXV * MAXV, INF);//初始化图G</span><br><span class="line">    for (int i = 0; i &lt; k; i++)&#123;</span><br><span class="line">        cin &gt;&gt; str1 &gt;&gt; str2;</span><br><span class="line">        int c1 = cityToIndex[str1], c2 = cityToIndex[str2];</span><br><span class="line">        cin &gt;&gt; G[c1][c2];</span><br><span class="line">        G[c2][c1] = G[c1][c2];//无向边</span><br><span class="line">    &#125;</span><br><span class="line">    Dijkstra(0);</span><br><span class="line">    int rom = cityToIndex[&quot;ROM&quot;];</span><br><span class="line">    printf(&quot;%d %d %d %d\n&quot;, num[rom], d[rom], w[rom], w[rom]/pt[rom]);</span><br><span class="line">    printPath(rom);</span><br><span class="line">    return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>
<p>dijstra + dfs:<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br><span class="line">73</span><br><span class="line">74</span><br><span class="line">75</span><br><span class="line">76</span><br><span class="line">77</span><br><span class="line">78</span><br><span class="line">79</span><br><span class="line">80</span><br><span class="line">81</span><br><span class="line">82</span><br><span class="line">83</span><br><span class="line">84</span><br><span class="line">85</span><br><span class="line">86</span><br><span class="line">87</span><br><span class="line">88</span><br><span class="line">89</span><br><span class="line">90</span><br><span class="line">91</span><br><span class="line">92</span><br><span class="line">93</span><br><span class="line">94</span><br><span class="line">95</span><br><span class="line">96</span><br><span class="line">97</span><br><span class="line">98</span><br><span class="line">99</span><br><span class="line">100</span><br><span class="line">101</span><br><span class="line">102</span><br><span class="line">103</span><br><span class="line">104</span><br><span class="line">105</span><br><span class="line">106</span><br><span class="line">107</span><br><span class="line">108</span><br><span class="line">109</span><br><span class="line">110</span><br><span class="line">111</span><br></pre></td><td class="code"><pre><span class="line">#include &quot;stdio.h&quot;</span><br><span class="line">#include &quot;math.h&quot;</span><br><span class="line">#include &quot;string.h&quot;</span><br><span class="line">#include &quot;iostream&quot;</span><br><span class="line">//#include &quot;stdlib.h&quot;</span><br><span class="line">#include &quot;vector&quot;</span><br><span class="line">#include &quot;set&quot;</span><br><span class="line">#include &quot;map&quot;</span><br><span class="line">//#include &quot;stack&quot;</span><br><span class="line">#include &quot;queue&quot;</span><br><span class="line">#include &quot;algorithm&quot;</span><br><span class="line">using namespace std;</span><br><span class="line"></span><br><span class="line">const int MAXV = 410;//最大顶点数</span><br><span class="line">const int INF = 1000000000;//无穷大</span><br><span class="line">//n 城市数目, k 边数, G 邻接矩阵, w 点权</span><br><span class="line">//d[]最短距离,numPath记录最短路径条数</span><br><span class="line">//maxW记录最大点权之和，maxAvg为最大平均点权</span><br><span class="line">int n, k, G[MAXV][MAXV], weight[MAXV];</span><br><span class="line">int d[MAXV], numPath = 0, maxW = 0;</span><br><span class="line">double maxAvg = 0;</span><br><span class="line">bool vis[MAXV] = &#123;false&#125;;</span><br><span class="line">vector&lt;int&gt; pre[MAXV];</span><br><span class="line">vector&lt;int&gt; tempPath, path;</span><br><span class="line">map&lt;string, int&gt; cityToIndex;</span><br><span class="line">map&lt;int, string&gt; indexToCity;</span><br><span class="line"></span><br><span class="line">void Dijstra(int s)&#123;//起点s</span><br><span class="line">    fill(d, d + MAXV, INF);</span><br><span class="line">    d[s] = 0;</span><br><span class="line">    for (int i = 0; i &lt; n; i++) &#123;</span><br><span class="line">        int u = -1, MIN = INF;</span><br><span class="line">        for (int j = 0; j &lt; n; j++) &#123;</span><br><span class="line">            if (vis[j] == false &amp;&amp; d[j] &lt; MIN) &#123;</span><br><span class="line">                u = j;</span><br><span class="line">                MIN = d[j];</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        if (u == -1) return;</span><br><span class="line">        vis[u] = true;</span><br><span class="line">        for (int v = 0; v &lt; n; v++) &#123;</span><br><span class="line">            if (vis[v] == false &amp;&amp; G[u][v] != INF) &#123;</span><br><span class="line">                if (d[u] + G[u][v] &lt; d[v]) &#123;</span><br><span class="line">                    d[v] = d[u] + G[u][v];</span><br><span class="line">                    pre[v].clear();</span><br><span class="line">                    pre[v].push_back(u);</span><br><span class="line">                &#125; else if(d[u] + G[u][v] == d[v])&#123;</span><br><span class="line">                    pre[v].push_back(u);</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">void DFS(int v)&#123;</span><br><span class="line">    if (v == 0) &#123;</span><br><span class="line">        tempPath.push_back(v);</span><br><span class="line">        numPath++;</span><br><span class="line">        int tempW = 0;//临时路径tempPath的点权之和</span><br><span class="line">        for (int i = tempPath.size() - 2; i &gt;= 0; i--) &#123;</span><br><span class="line">            int id = tempPath[i];</span><br><span class="line">            tempW += weight[id];</span><br><span class="line">        &#125;</span><br><span class="line">        double tempAvg = 1.0 * tempW / (tempPath.size() - 1);</span><br><span class="line">        if (tempW &gt; maxW) &#123;//当前点权和更大</span><br><span class="line">            maxW = tempW;</span><br><span class="line">            maxAvg = tempAvg;</span><br><span class="line">            path = tempPath;</span><br><span class="line">        &#125;else if (tempW == maxAvg &amp;&amp; tempAvg &gt; maxAvg)&#123;</span><br><span class="line">            maxAvg = tempAvg;</span><br><span class="line">            path = tempPath;</span><br><span class="line">        &#125;</span><br><span class="line">        tempPath.pop_back();</span><br><span class="line">        return;</span><br><span class="line">    &#125;</span><br><span class="line">    tempPath.push_back(v);</span><br><span class="line">    for (int i = 0; i &lt; pre[v].size(); i++) &#123;</span><br><span class="line">        DFS(pre[v][i]);</span><br><span class="line">    &#125;</span><br><span class="line">    tempPath.pop_back();</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">int main()&#123;</span><br><span class="line">    string start, city1, city2;</span><br><span class="line">    cin &gt;&gt; n &gt;&gt; k &gt;&gt; start;</span><br><span class="line">    cityToIndex[start] = 0;</span><br><span class="line">    indexToCity[0] = start;</span><br><span class="line">    for (int i = 1; i &lt;= n -1; i++) &#123;</span><br><span class="line">        cin &gt;&gt; city1 &gt;&gt; weight[i];</span><br><span class="line">        cityToIndex[city1] = i;</span><br><span class="line">        indexToCity[i] = city1;</span><br><span class="line">    &#125;</span><br><span class="line">    fill(G[0], G[0] + MAXV * MAXV, INF);</span><br><span class="line">    for (int i = 0; i &lt; k; i++) &#123;</span><br><span class="line">        cin &gt;&gt; city1 &gt;&gt; city2;</span><br><span class="line">        int c1 = cityToIndex[city1], c2 = cityToIndex[city2];</span><br><span class="line">        cin &gt;&gt; G[c1][c2];</span><br><span class="line">        G[c2][c1] = G[c1][c2];</span><br><span class="line">    &#125;</span><br><span class="line">    Dijstra(0);</span><br><span class="line">    int rom = cityToIndex[&quot;ROM&quot;];</span><br><span class="line">    DFS(rom);</span><br><span class="line">    printf(&quot;%d %d %d %d\n&quot;, numPath, d[rom], maxW, (int)maxAvg);</span><br><span class="line">    for (int i = path.size() - 1; i &gt;= 0; i--) &#123;</span><br><span class="line">        cout &lt;&lt; indexToCity[path[i]];</span><br><span class="line">        if (i &gt; 0) &#123;</span><br><span class="line">            cout &lt;&lt;&quot;-&gt;&quot;;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>

          
        
      
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            <p>A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. However it must guarantee that all the houses are in its service range.</p>
<p>Now given the map of the city and several candidate locations for the gas station, you are supposed to give the best recommendation. If there are more than one solution, output the one with the smallest average distance to all the houses. If such a solution is still not unique, output the one with the smallest index number.</p>
<p>Input Specification:</p>
<p>Each input file contains one test case. For each case, the first line contains 4 positive integers: N (&lt;= 103), the total number of houses; M (&lt;= 10), the total number of the candidate locations for the gas stations; K (&lt;= 104), the number of roads connecting the houses and the gas stations; and DS, the maximum service range of the gas station. It is hence assumed that all the houses are numbered from 1 to N, and all the candidate locations are numbered from G1 to GM.</p>
<p>Then K lines follow, each describes a road in the format<br>P1 P2 Dist<br>where P1 and P2 are the two ends of a road which can be either house numbers or gas station numbers, and Dist is the integer length of the road.</p>
<p>Output Specification:</p>
<p>For each test case, print in the first line the index number of the best location. In the next line, print the minimum and the average distances between the solution and all the houses. The numbers in a line must be separated by a space and be accurate up to 1 decimal place. If the solution does not exist, simply output “No Solution”.</p>
<p>Sample Input 1:<br>4 3 11 5<br>1 2 2<br>1 4 2<br>1 G1 4<br>1 G2 3<br>2 3 2<br>2 G2 1<br>3 4 2<br>3 G3 2<br>4 G1 3<br>G2 G1 1<br>G3 G2 2<br>Sample Output 1:<br>G1<br>2.0 3.3<br>Sample Input 2:<br>2 1 2 10<br>1 G1 9<br>2 G1 20<br>Sample Output 2:<br>No Solution<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br><span class="line">73</span><br><span class="line">74</span><br><span class="line">75</span><br><span class="line">76</span><br><span class="line">77</span><br><span class="line">78</span><br><span class="line">79</span><br><span class="line">80</span><br><span class="line">81</span><br><span class="line">82</span><br><span class="line">83</span><br><span class="line">84</span><br><span class="line">85</span><br><span class="line">86</span><br><span class="line">87</span><br><span class="line">88</span><br><span class="line">89</span><br><span class="line">90</span><br><span class="line">91</span><br><span class="line">92</span><br><span class="line">93</span><br><span class="line">94</span><br><span class="line">95</span><br><span class="line">96</span><br><span class="line">97</span><br><span class="line">98</span><br><span class="line">99</span><br><span class="line">100</span><br><span class="line">101</span><br><span class="line">102</span><br><span class="line">103</span><br><span class="line">104</span><br><span class="line">105</span><br></pre></td><td class="code"><pre><span class="line">#include &quot;stdio.h&quot;</span><br><span class="line">#include &quot;math.h&quot;</span><br><span class="line">#include &quot;string.h&quot;</span><br><span class="line">#include &quot;iostream&quot;</span><br><span class="line">//#include &quot;stdlib.h&quot;</span><br><span class="line">#include &quot;vector&quot;</span><br><span class="line">#include &quot;set&quot;</span><br><span class="line">#include &quot;map&quot;</span><br><span class="line">//#include &quot;stack&quot;</span><br><span class="line">#include &quot;queue&quot;</span><br><span class="line">#include &quot;algorithm&quot;</span><br><span class="line">using namespace std;</span><br><span class="line"></span><br><span class="line">const int MAXV = 1020;//最大顶点数</span><br><span class="line">const int INF = 1000000000;//无穷大</span><br><span class="line"></span><br><span class="line">//n为顶点数，m为加油站数，k为边数，DS为服务范围，G为邻接矩阵</span><br><span class="line">//d[]记录最短距离</span><br><span class="line">int n, m, k, DS, G[MAXV][MAXV];</span><br><span class="line">int d[MAXV];</span><br><span class="line">bool vis[MAXV] = &#123;false&#125;;</span><br><span class="line"></span><br><span class="line">//Dijkstra算法求所有顶点到起点s的最短距离</span><br><span class="line">void Dijkstra(int s)&#123;</span><br><span class="line">    memset(vis, false, sizeof(vis));</span><br><span class="line">    fill(d, d + MAXV, INF);</span><br><span class="line">    d[s] = 0;</span><br><span class="line">    for (int i = 0; i &lt; n + m; i++) &#123;</span><br><span class="line">        int u = -1, MIN = INF;</span><br><span class="line">        for (int j = 1; j &lt;= n + m; j++) &#123;</span><br><span class="line">            if (vis[j] == false &amp;&amp; d[j] &lt; MIN) &#123;</span><br><span class="line">                u = j;</span><br><span class="line">                MIN = d[j];</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        if (u == -1) return;</span><br><span class="line">        vis[u] = true;</span><br><span class="line">        for (int v = 1; v &lt;= n + m; v++) &#123;</span><br><span class="line">            if (vis[v] == false &amp;&amp; G[u][v] != INF) &#123;</span><br><span class="line">                if (d[u] + G[u][v] &lt; d[v]) &#123;</span><br><span class="line">                    d[v] = d[u] + G[u][v];//覆盖d[v]</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br><span class="line">//将str[]转换为数字，若str是数字，则返回本身；否则返回去掉G之后的数加上n</span><br><span class="line">int getID(char str[])&#123;</span><br><span class="line">    int i = 0, len = strlen(str), ID = 0;</span><br><span class="line">    while (i &lt; len) &#123;</span><br><span class="line">        if (str[i] != &apos;G&apos;) &#123;</span><br><span class="line">            ID = ID * 10 + (str[i] - &apos;0&apos;);</span><br><span class="line">        &#125;</span><br><span class="line">        i++;</span><br><span class="line">    &#125;</span><br><span class="line">    if (str[0] == &apos;G&apos;) &#123;</span><br><span class="line">        return ID + n;</span><br><span class="line">    &#125;else&#123;</span><br><span class="line">        return ID;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">int main()&#123;</span><br><span class="line">    scanf(&quot;%d%d%d%d&quot;, &amp;n, &amp;m, &amp;k, &amp;DS);</span><br><span class="line">    int u, v, w;</span><br><span class="line">    char city1[5], city2[5];</span><br><span class="line">    fill(G[0], G[0] + MAXV * MAXV, INF);</span><br><span class="line">    for (int i = 0; i &lt; k; i++) &#123;</span><br><span class="line">        scanf(&quot;%s %s %d&quot;, city1, city2, &amp;w);</span><br><span class="line">        u = getID(city1);</span><br><span class="line">        v = getID(city2);</span><br><span class="line">        G[u][v] = G[v][u] = w;//边权</span><br><span class="line">    &#125;</span><br><span class="line">    //ansDis存放最大的最短距离</span><br><span class="line">    //ansAvg存放最小平均距离，ansID存放最终加油站ID</span><br><span class="line">    double ansDis = -1, ansAvg = INF;</span><br><span class="line">    int ansID = -1;</span><br><span class="line">    for (int i = n + 1; i &lt;= n + m; i++) &#123;//枚举所有加油站</span><br><span class="line">        double minDis = INF, avg = 0;//minDis为最大的最近距离，avg为平均距离</span><br><span class="line">        Dijkstra(i);//求出d数组</span><br><span class="line">        for (int j = 1; j &lt;= n; j++) &#123;//枚举所有民房，求出minDis与avg</span><br><span class="line">            if (d[j] &gt; DS) &#123;//存在距离大于DS的居民房，跳出</span><br><span class="line">                minDis = -1;</span><br><span class="line">                break;</span><br><span class="line">            &#125;</span><br><span class="line">            if (d[j] &lt; minDis) minDis = d[j];</span><br><span class="line">            avg += 1.0 * d[j] / n;//获取平均距离</span><br><span class="line">        &#125;</span><br><span class="line">        if (minDis == -1) continue;//跳过这个加油站</span><br><span class="line">        if (minDis &gt; ansDis) &#123;//最大距离</span><br><span class="line">            ansDis = minDis;</span><br><span class="line">            ansID = i;</span><br><span class="line">            ansAvg = avg;</span><br><span class="line">        &#125;else if (minDis == ansDis &amp;&amp; avg &lt; ansAvg)&#123;//更新最小平均距离</span><br><span class="line">            ansID = i;</span><br><span class="line">            ansAvg = avg;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    if (ansID == -1) printf(&quot;No Solution\n&quot;);//无解</span><br><span class="line">    else&#123;</span><br><span class="line">        printf(&quot;G%d\n&quot;, ansID - n);</span><br><span class="line">        printf(&quot;%.1f %.1f\n&quot;, ansDis, ansAvg);</span><br><span class="line">    &#125;</span><br><span class="line">    return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>

          
        
      
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                <a class="post-title-link" href="/2017/02/26/233/" itemprop="url">PAT A1030</a></h1>
        

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            <p>A traveler’s map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination. If such a shortest path is not unique, you are supposed to output the one with the minimum cost, which is guaranteed to be unique.</p>
<p>Input Specification:</p>
<p>Each input file contains one test case. Each case starts with a line containing 4 positive integers N, M, S, and D, where N (&lt;=500) is the number of cities (and hence the cities are numbered from 0 to N-1); M is the number of highways; S and D are the starting and the destination cities, respectively. Then M lines follow, each provides the information of a highway, in the format:</p>
<p>City1 City2 Distance Cost</p>
<p>where the numbers are all integers no more than 500, and are separated by a space.</p>
<p>Output Specification:</p>
<p>For each test case, print in one line the cities along the shortest path from the starting point to the destination, followed by the total distance and the total cost of the path. The numbers must be separated by a space and there must be no extra space at the end of output.</p>
<p>Sample Input<br>4 5 0 3<br>0 1 1 20<br>1 3 2 30<br>0 3 4 10<br>0 2 2 20<br>2 3 1 20<br>Sample Output<br>0 2 3 3 40<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br><span class="line">73</span><br><span class="line">74</span><br><span class="line">75</span><br><span class="line">76</span><br><span class="line">77</span><br></pre></td><td class="code"><pre><span class="line">#include &quot;stdio.h&quot;</span><br><span class="line">#include &quot;math.h&quot;</span><br><span class="line">#include &quot;string.h&quot;</span><br><span class="line">#include &quot;iostream&quot;</span><br><span class="line">//#include &quot;stdlib.h&quot;</span><br><span class="line">#include &quot;vector&quot;</span><br><span class="line">#include &quot;set&quot;</span><br><span class="line">#include &quot;map&quot;</span><br><span class="line">//#include &quot;stack&quot;</span><br><span class="line">#include &quot;queue&quot;</span><br><span class="line">#include &quot;algorithm&quot;</span><br><span class="line">using namespace std;</span><br><span class="line"></span><br><span class="line">const int MAXV = 510;//最大顶点数</span><br><span class="line">const int INF = 1000000000;//无穷大</span><br><span class="line"></span><br><span class="line">//n为顶点数，m为边数，st和ed为起点和终点</span><br><span class="line">//G为距离矩阵，cost为花费矩阵</span><br><span class="line">//d[]记录最短距离，c[]记录最小花费</span><br><span class="line">int n, m, st, ed, G[MAXV][MAXV], cost[MAXV][MAXV];</span><br><span class="line">int d[MAXV], c[MAXV], pre[MAXV];</span><br><span class="line">bool vis[MAXV] = &#123;false&#125;;//vis[i]表示顶点i已访问，初值均为false</span><br><span class="line"></span><br><span class="line">void Dijkstra(int s)&#123;//s为起点</span><br><span class="line">    fill(d, d + MAXV, INF);//fill函数将整个d数组赋为INF</span><br><span class="line">    fill(c, c + MAXV, INF);//fill函数将整个c数组赋为INF</span><br><span class="line">    d[s] = 0;</span><br><span class="line">    c[s] = 0;</span><br><span class="line">    for (int i = 0; i &lt; n; i++) &#123;</span><br><span class="line">        int u = -1, MIN = INF;</span><br><span class="line">        for (int j = 0; j &lt; n; j++) &#123;</span><br><span class="line">            if (vis[j] == false &amp;&amp; d[j] &lt; MIN) &#123;</span><br><span class="line">                MIN = d[j];</span><br><span class="line">                u = j;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        if (u == -1) return;</span><br><span class="line">        vis[u] = true;</span><br><span class="line">        for (int v = 0; v &lt; n; v++) &#123;</span><br><span class="line">            if (vis[v] == false &amp;&amp; G[u][v] != INF) &#123;</span><br><span class="line">                if (d[u] +  G[u][v] &lt; d[v]) &#123;</span><br><span class="line">                    d[v] = d[u] + G[u][v];</span><br><span class="line">                    c[v] = c[u] + cost[u][v];</span><br><span class="line">                    pre[v] = u;</span><br><span class="line">                &#125;else if(d[u] + G[u][v] == d[v] &amp;&amp; c[u] + cost[u][v] &lt; c[v])&#123;</span><br><span class="line">                    c[v] = c[u] + cost[u][v];</span><br><span class="line">                    pre[v] = u;</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">void DFS(int v)&#123;//打印路径</span><br><span class="line">    if (v == st) &#123;//递归终点</span><br><span class="line">        printf(&quot;%d &quot;, v);</span><br><span class="line">        return;</span><br><span class="line">    &#125;</span><br><span class="line">    DFS(pre[v]);</span><br><span class="line">    printf(&quot;%d &quot;, v);</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">int main()&#123;</span><br><span class="line">    scanf(&quot;%d%d%d%d&quot;, &amp;n, &amp;m, &amp;st, &amp;ed);</span><br><span class="line">    int u, v;</span><br><span class="line">    fill(G[0], G[0] + MAXV * MAXV, INF);//初始化图</span><br><span class="line">    for (int i = 0; i &lt; m; i++) &#123;</span><br><span class="line">        scanf(&quot;%d%d&quot;, &amp;u, &amp;v);</span><br><span class="line">        scanf(&quot;%d%d&quot;, &amp;G[u][v], &amp;cost[u][v]);</span><br><span class="line">        G[v][u] = G[u][v];</span><br><span class="line">        cost[v][u] = cost[u][v];</span><br><span class="line">    &#125;</span><br><span class="line">    Dijkstra(st);</span><br><span class="line">    DFS(ed);</span><br><span class="line">    printf(&quot;%d %d\n&quot;, d[ed], c[ed]);</span><br><span class="line">    return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>
<p>dijkstra + dfs做法<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br><span class="line">73</span><br><span class="line">74</span><br><span class="line">75</span><br><span class="line">76</span><br><span class="line">77</span><br><span class="line">78</span><br><span class="line">79</span><br><span class="line">80</span><br><span class="line">81</span><br><span class="line">82</span><br><span class="line">83</span><br><span class="line">84</span><br><span class="line">85</span><br><span class="line">86</span><br><span class="line">87</span><br><span class="line">88</span><br><span class="line">89</span><br><span class="line">90</span><br><span class="line">91</span><br><span class="line">92</span><br><span class="line">93</span><br></pre></td><td class="code"><pre><span class="line">#include &quot;stdio.h&quot;</span><br><span class="line">#include &quot;math.h&quot;</span><br><span class="line">#include &quot;string.h&quot;</span><br><span class="line">#include &quot;iostream&quot;</span><br><span class="line">//#include &quot;stdlib.h&quot;</span><br><span class="line">#include &quot;vector&quot;</span><br><span class="line">#include &quot;set&quot;</span><br><span class="line">#include &quot;map&quot;</span><br><span class="line">//#include &quot;stack&quot;</span><br><span class="line">#include &quot;queue&quot;</span><br><span class="line">#include &quot;algorithm&quot;</span><br><span class="line">using namespace std;</span><br><span class="line"></span><br><span class="line">const int MAXV = 510;//最大顶点数</span><br><span class="line">const int INF = 1000000000;//无穷大</span><br><span class="line"></span><br><span class="line">//n为顶点数，m为边数，st和ed为起点和终点</span><br><span class="line">//G为距离矩阵，cost为花费矩阵</span><br><span class="line">//d[]记录最短距离，c[]记录最小花费</span><br><span class="line">int n, m, st, ed, G[MAXV][MAXV], cost[MAXV][MAXV];</span><br><span class="line">int d[MAXV], minCost = INF;</span><br><span class="line">bool vis[MAXV] = &#123;false&#125;;//vis[i]表示顶点i已访问，初值均为false</span><br><span class="line">vector&lt;int&gt; pre[MAXV];//前驱</span><br><span class="line">vector&lt;int&gt; tempPath, Path;//临时路径，最优路径</span><br><span class="line"></span><br><span class="line">void Dijkstra(int s)&#123;//s为起点</span><br><span class="line">    fill(d, d + MAXV, INF);//fill函数将整个d数组赋为INF</span><br><span class="line">    d[s] = 0;</span><br><span class="line">    for (int i = 0; i &lt; n; i++) &#123;</span><br><span class="line">        int u = -1, MIN = INF;</span><br><span class="line">        for (int j = 0; j &lt; n; j++) &#123;</span><br><span class="line">            if (vis[j] == false &amp;&amp; d[j] &lt; MIN) &#123;</span><br><span class="line">                MIN = d[j];</span><br><span class="line">                u = j;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        if (u == -1) return;</span><br><span class="line">        vis[u] = true;</span><br><span class="line">        for (int v = 0; v &lt; n; v++) &#123;</span><br><span class="line">            if (vis[v] == false &amp;&amp; G[u][v] != INF) &#123;</span><br><span class="line">                if (d[u] +  G[u][v] &lt; d[v]) &#123;</span><br><span class="line">                    d[v] = d[u] + G[u][v];</span><br><span class="line">                    pre[v].clear();</span><br><span class="line">                    pre[v].push_back(u);</span><br><span class="line">                &#125;else if(d[u] + G[u][v] == d[v])&#123;</span><br><span class="line">                    pre[v].push_back(u);</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">void DFS(int v)&#123;//打印路径</span><br><span class="line">    if (v == st) &#123;//递归边界</span><br><span class="line">        tempPath.push_back(v);//加入st</span><br><span class="line">        int tempCost = 0;//记录当前路径花费之和(边权和)</span><br><span class="line">        for (int i = tempPath.size() - 1; i &gt; 0; i--) &#123;</span><br><span class="line">            //id当前, idNext下一个id</span><br><span class="line">            int id = tempPath[i], idNext = tempPath[i - 1];</span><br><span class="line">            tempCost += cost[id][idNext];</span><br><span class="line">        &#125;</span><br><span class="line">        if (tempCost &lt; minCost) &#123;</span><br><span class="line">            minCost = tempCost;</span><br><span class="line">            Path = tempPath;</span><br><span class="line">        &#125;</span><br><span class="line">        tempPath.pop_back();//弹出st</span><br><span class="line">        return;</span><br><span class="line">    &#125;</span><br><span class="line">    tempPath.push_back(v);</span><br><span class="line">    for (int i = 0; i &lt; pre[v].size(); i++) &#123;</span><br><span class="line">        DFS(pre[v][i]);</span><br><span class="line">    &#125;</span><br><span class="line">    tempPath.pop_back();</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">int main()&#123;</span><br><span class="line">    scanf(&quot;%d%d%d%d&quot;, &amp;n, &amp;m, &amp;st, &amp;ed);</span><br><span class="line">    int u, v;</span><br><span class="line">    fill(G[0], G[0] + MAXV * MAXV, INF);//初始化图</span><br><span class="line">    for (int i = 0; i &lt; m; i++) &#123;</span><br><span class="line">        scanf(&quot;%d%d&quot;, &amp;u, &amp;v);</span><br><span class="line">        scanf(&quot;%d%d&quot;, &amp;G[u][v], &amp;cost[u][v]);</span><br><span class="line">        G[v][u] = G[u][v];</span><br><span class="line">        cost[v][u] = cost[u][v];</span><br><span class="line">    &#125;</span><br><span class="line">    Dijkstra(st);</span><br><span class="line">    DFS(ed);</span><br><span class="line">    for (int i = Path.size() - 1; i &gt;= 0; i--) &#123;</span><br><span class="line">        printf(&quot;%d &quot;, Path[i]);//打印路径</span><br><span class="line">    &#125;</span><br><span class="line">    printf(&quot;%d %d\n&quot;, d[ed], minCost);</span><br><span class="line">    return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>

          
        
      
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                <a class="post-title-link" href="/2017/02/25/231/" itemprop="url">PAT A1018</a></h1>
        

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            <p>There is a public bike service in Hangzhou City which provides great convenience to the tourists from all over the world. One may rent a bike at any station and return it to any other stations in the city.</p>
<p>The Public Bike Management Center (PBMC) keeps monitoring the real-time capacity of all the stations. A station is said to be in perfect condition if it is exactly half-full. If a station is full or empty, PBMC will collect or send bikes to adjust the condition of that station to perfect. And more, all the stations on the way will be adjusted as well.</p>
<p>When a problem station is reported, PBMC will always choose the shortest path to reach that station. If there are more than one shortest path, the one that requires the least number of bikes sent from PBMC will be chosen.<br><img src="https://www.patest.cn/upload/11_lrrk0m5o4kg.jpg" alt="请输入图片描述"></p>
<p>Figure 1<br>Figure 1 illustrates an example. The stations are represented by vertices and the roads correspond to the edges. The number on an edge is the time taken to reach one end station from another. The number written inside a vertex S is the current number of bikes stored at S. Given that the maximum capacity of each station is 10. To solve the problem at S3, we have 2 different shortest paths:</p>
<ol>
<li><p>PBMC -&gt; S1 -&gt; S3. In this case, 4 bikes must be sent from PBMC, because we can collect 1 bike from S1 and then take 5 bikes to S3, so that both stations will be in perfect conditions.</p>
</li>
<li><p>PBMC -&gt; S2 -&gt; S3. This path requires the same time as path 1, but only 3 bikes sent from PBMC and hence is the one that will be chosen.</p>
</li>
</ol>
<p>Input Specification:</p>
<p>Each input file contains one test case. For each case, the first line contains 4 numbers: Cmax (&lt;= 100), always an even number, is the maximum capacity of each station; N (&lt;= 500), the total number of stations; Sp, the index of the problem station (the stations are numbered from 1 to N, and PBMC is represented by the vertex 0); and M, the number of roads. The second line contains N non-negative numbers Ci (i=1,…N) where each Ci is the current number of bikes at Si respectively. Then M lines follow, each contains 3 numbers: Si, Sj, and Tij which describe the time Tij taken to move betwen stations Si and Sj. All the numbers in a line are separated by a space.</p>
<p>Output Specification:</p>
<p>For each test case, print your results in one line. First output the number of bikes that PBMC must send. Then after one space, output the path in the format: 0-&gt;S1-&gt;…-&gt;Sp. Finally after another space, output the number of bikes that we must take back to PBMC after the condition of Sp is adjusted to perfect.</p>
<p>Note that if such a path is not unique, output the one that requires minimum number of bikes that we must take back to PBMC. The judge’s data guarantee that such a path is unique.</p>
<p>Sample Input:<br>10 3 3 5<br>6 7 0<br>0 1 1<br>0 2 1<br>0 3 3<br>1 3 1<br>2 3 1<br>Sample Output:<br>3 0-&gt;2-&gt;3 0</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br><span class="line">73</span><br><span class="line">74</span><br><span class="line">75</span><br><span class="line">76</span><br><span class="line">77</span><br><span class="line">78</span><br><span class="line">79</span><br><span class="line">80</span><br><span class="line">81</span><br><span class="line">82</span><br><span class="line">83</span><br><span class="line">84</span><br><span class="line">85</span><br><span class="line">86</span><br><span class="line">87</span><br><span class="line">88</span><br><span class="line">89</span><br><span class="line">90</span><br><span class="line">91</span><br><span class="line">92</span><br><span class="line">93</span><br><span class="line">94</span><br><span class="line">95</span><br><span class="line">96</span><br><span class="line">97</span><br><span class="line">98</span><br><span class="line">99</span><br><span class="line">100</span><br><span class="line">101</span><br><span class="line">102</span><br><span class="line">103</span><br><span class="line">104</span><br><span class="line">105</span><br><span class="line">106</span><br><span class="line">107</span><br><span class="line">108</span><br><span class="line">109</span><br><span class="line">110</span><br><span class="line">111</span><br><span class="line">112</span><br><span class="line">113</span><br><span class="line">114</span><br><span class="line">115</span><br><span class="line">116</span><br></pre></td><td class="code"><pre><span class="line">#include &quot;stdio.h&quot;</span><br><span class="line">#include &quot;math.h&quot;</span><br><span class="line">#include &quot;string.h&quot;</span><br><span class="line">#include &quot;iostream&quot;</span><br><span class="line">//#include &quot;stdlib.h&quot;</span><br><span class="line">#include &quot;vector&quot;</span><br><span class="line">#include &quot;set&quot;</span><br><span class="line">#include &quot;map&quot;</span><br><span class="line">//#include &quot;stack&quot;</span><br><span class="line">#include &quot;queue&quot;</span><br><span class="line">#include &quot;algorithm&quot;</span><br><span class="line">using namespace std;</span><br><span class="line"></span><br><span class="line">const int MAXV = 510;//最大顶点数</span><br><span class="line">const int INF = 1000000000;//无穷大</span><br><span class="line"></span><br><span class="line">//n为顶点数，m为边数，Cmax为最大容量，Sp为问题站点</span><br><span class="line">//G为邻接矩阵，weight为点权，d[]记录最短距离</span><br><span class="line">//mindNeed记录最少携带数目，minRemain记录最少带回的数目</span><br><span class="line">int n, m , Cmax, Sp, numPath = 0, G[MAXV][MAXV], weight[MAXV];</span><br><span class="line">int d[MAXV], minNeed = INF, minRemain = INF;</span><br><span class="line">bool vis[MAXV] = &#123;false&#125;;//vis[i] == true 表示顶点i已访问，初始均为false</span><br><span class="line">vector&lt;int&gt; pre[MAXV];//前驱</span><br><span class="line">vector&lt;int&gt; tempPath, path;//临时路径及最优路径</span><br><span class="line"></span><br><span class="line">void Dijkstra(int s)&#123;//s为起点</span><br><span class="line">    fill(d, d + MAXV, INF);</span><br><span class="line">    d[s] = 0;</span><br><span class="line">    for (int i = 0; i &lt; n; i++) &#123;//循环n次</span><br><span class="line">        int u = -1, MIN = INF;//u使d[u]最小，MIN存放该最小的d[u]</span><br><span class="line">        for (int j = 0; j &lt;= n; j++) &#123;//找到未访问的顶点中d[]最小的</span><br><span class="line">            if (vis[j] == false &amp;&amp; d[j] &lt; MIN) &#123;</span><br><span class="line">                u = j;</span><br><span class="line">                MIN = d[j];</span><br><span class="line">            &#125;</span><br><span class="line">            </span><br><span class="line">        &#125;</span><br><span class="line">        //找不到小于INF的d[u]，说明剩下的顶点和起点s不连通</span><br><span class="line">        if (u == -1) return;</span><br><span class="line">        vis[u] = true;//标记u已访问</span><br><span class="line">        for (int v = 0; v &lt;= n; v++) &#123;</span><br><span class="line">            //如果v未访问&amp;&amp; u能到达v</span><br><span class="line">            if (vis[v] == false &amp;&amp; G[u][v] != INF) &#123;</span><br><span class="line">                if (d[u] + G[u][v] &lt; d[v]) &#123;</span><br><span class="line">                    d[v] = d[u] + G[u][v];//优化d[v]</span><br><span class="line">                    pre[v].clear();</span><br><span class="line">                    pre[v].push_back(u);</span><br><span class="line">                &#125;else if (d[u] + G[u][v] == d[v])&#123;</span><br><span class="line">                    pre[v].push_back(u);</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br><span class="line">void DFS(int v)&#123;</span><br><span class="line">    if(v == 0)&#123;//递归边界,叶子结点</span><br><span class="line">        tempPath.push_back(v);</span><br><span class="line">        //路径tempPath上需要携带数目、需要带回的数目</span><br><span class="line">        int need = 0, remain = 0;</span><br><span class="line">        for (int i = tempPath.size() - 1; i &gt;= 0 ; i--) &#123;</span><br><span class="line">            int id = tempPath[i];//当前结点编号id</span><br><span class="line">            if (weight[id] &gt; 0) &#123;//点权大于0，说明需要带走一部分自行车</span><br><span class="line">                remain += weight[id];</span><br><span class="line">            &#125;else&#123;//点权不超过0，需要补给</span><br><span class="line">                if (remain &gt; abs(weight[id])) &#123;//当前持有量足够补给</span><br><span class="line">                    remain -= abs(weight[id]);</span><br><span class="line">                &#125;else&#123;</span><br><span class="line">                    need += abs(weight[id]) - remain;//不够的部分从PBMC携带</span><br><span class="line">                    remain = 0;//当前持有的自行车都用来补给</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        if (need &lt; minNeed)&#123;//需要从PBMC携带的自行车数目更少</span><br><span class="line">            minNeed = need;//优化minNeed</span><br><span class="line">            minRemain = remain;//覆盖minRemain</span><br><span class="line">            path = tempPath;//覆盖最优路径path</span><br><span class="line">        &#125;else if(need == minNeed &amp;&amp; remain &lt; minRemain)&#123;</span><br><span class="line">            //携带数目相同，带回数目变少</span><br><span class="line">            minRemain = remain;//优化minRemain</span><br><span class="line">            path = tempPath;//覆盖最优路径path</span><br><span class="line">        &#125;</span><br><span class="line">        tempPath.pop_back();</span><br><span class="line">        return;</span><br><span class="line">    &#125;</span><br><span class="line">    tempPath.push_back(v);</span><br><span class="line">    for (int i = 0; i &lt; pre[v].size(); i++) &#123;</span><br><span class="line">        DFS(pre[v][i]);</span><br><span class="line">    &#125;</span><br><span class="line">    tempPath.pop_back();</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">int main()&#123;</span><br><span class="line">    scanf(&quot;%d%d%d%d&quot;, &amp;Cmax, &amp;n, &amp;Sp, &amp;m);</span><br><span class="line">    int u, v;</span><br><span class="line">    fill(G[0], G[0] + MAXV * MAXV, INF);</span><br><span class="line">    for (int i = 1; i &lt;= n; i++) &#123;</span><br><span class="line">        scanf(&quot;%d&quot;, &amp;weight[i]);</span><br><span class="line">        weight[i] -= Cmax / 2;//点权减去答案的一半</span><br><span class="line">    &#125;</span><br><span class="line">    for (int i = 0; i &lt; m; i++) &#123;</span><br><span class="line">        scanf(&quot;%d%d&quot;, &amp;u, &amp;v);</span><br><span class="line">        scanf(&quot;%d&quot;, &amp;G[u][v]);</span><br><span class="line">        G[v][u] = G[u][v];</span><br><span class="line">    &#125;</span><br><span class="line">    Dijkstra(0);//Dijkstra算法入口</span><br><span class="line">    DFS(Sp);</span><br><span class="line">    printf(&quot;%d &quot;, minNeed);</span><br><span class="line">    for (int i = path.size() - 1; i &gt;= 0; i--) &#123;</span><br><span class="line">        printf(&quot;%d&quot;, path[i]);</span><br><span class="line">        if (i &gt; 0) &#123;</span><br><span class="line">            printf(&quot;-&gt;&quot;);</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    printf(&quot; %d\n&quot;, minRemain);</span><br><span class="line">    return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

          
        
      
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